What is the Sum of the Polynomials
Polynomials: Sums and Products of Roots
Roots of a Polynomial
A “root” (or “zero”) is where the polynomial
is equal to zero:
Put but: a root is the xvalue where the yvalue equals zero.
General Polynomial
If we have a full general polynomial similar this:
f(x) = ax^{due north}
+ bx^{north1}
+ cx^{due north2}
+ … + z
Then:

Adding
the roots gives
−b/a 
Multiplying
the roots gives:
z/a
(for fiftyfifty degree polynomials similar quadratics) 
−z/a
(for odd degree polynomials like cubics)

z/a
Which can sometimes assistance us solve things.
How does this magic work? Let’s find out …
Factors
We can take a polynomial, such as:
f(x) = ax^{northward}
+ bx^{n1}
+ cx^{n2}
+ … + z
And so factor it like this:
f(x) = a(x−p)(ten−q)(ten−r)…
And then p, q, r, etc are the
roots
(where the polynomial equals zero)
Quadratic
Let’due south endeavour this with a Quadratic (where the variable’s biggest exponent is 2):
ax^{2}
+ bx + c
When the roots are
p
and
q, the aforementioned quadratic becomes:
a(x−p)(10−q)
Is there a relationship between
a,b,c
and
p,q
?
Let’south expand
a(10−p)(x−q):
a(x−p)(x−q)
= a( ten^{2}
− px − qx + pq )
= ax^{2}
− a(p+q)ten + apq
Now let united states of america compare:
Quadratic: 
ax^{2} 
+bx  +c 
Expanded Factors: 
ax^{2} 
−a(p+q)x  +apq 
We tin now come across that
−a(p+q)x = bx, then:
−a(p+q) = b
p+q = −b/a
And
apq
= c, so:
pq = c/a
And we get this result:
 Adding the roots gives
−b/a  Multiplying the roots gives
c/a
This can aid u.s. answer questions.
Example: What is an equation whose roots are five + √ii and 5 − √2
The sum of the roots is (5 + √2) + (5 − √two) =
x
The product of the roots is (5 + √ii) (5 − √2) = 25 − two =
23
And we desire an equation like:
ax^{2}
+ bx + c = 0
When
a=1
we can work out that:
 Sum of the roots =
−b/a
=
b  Product of the roots =
c/a
=
c
Which gives us this result
ten^{ii}
− (sum of the roots)x + (production of the roots) = 0
The sum of the roots is 10, and product of the roots is 23, so we go:
x^{two}
− 10x + 23 = 0
And here is its plot:
(Question: what happens if we cull
a=−1
?)
Cubic
Now let the states look at a Cubic (i caste higher than Quadratic):
ax^{3}
+ bx^{2}
+ cx + d
As with the Quadratic, let us expand the factors:
a(ten−p)(x−q)(10−r)
= ax^{three}
− a(p+q+r)x^{ii}
+ a(pq+pr+qr)x − a(pqr)
And we become:
Cubic: 
ax^{3} 
+bx^{2} 
+cx  +d 
Expanded Factors: 
ax^{three} 
−a(p+q+r)x^{two} 
+a(pq+pr+qr)x  −apqr 
We tin at present see that
−a(p+q+r)x^{2}
= bx^{two}
, so:
−a(p+q+r) = b
p+q+r = −b/a
And
−apqr
= d, then:
pqr = −d/a
This is interesting … nosotros get the same sort of thing:
 Calculation the roots gives
−b/a
(exactly the aforementioned every bit the Quadratic)  Multiplying the roots gives
−d/a
(like to +c/a for the Quadratic)
(We also become
pq+pr+qr = c/a, which can itself exist useful.)
Higher Polynomials
The same pattern continues with higher polynomials.
In General:
 Adding the roots gives
−b/a  Multiplying the roots gives (where “z” is the constant at the terminate):

z/a
(for even degree polynomials like quadratics) 
−z/a
(for odd degree polynomials like cubics)

z/a
What is the Sum of the Polynomials
Source: https://www.mathsisfun.com/algebra/polynomialssumsproductsroots.html